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Question:

The motion of a body in simple harmonic motion is given by the displacement function,

x (t) = A cos (ωt + φ).

Given that at t = 0, the initial velocity of the body is ω cm/s and its initial position is 1 cm, calculate its initial phase angle and amplitude?

If in place of the cosine function, a sine function is used to represent the simple harmonic motion:

x = B sin (ωt + α),

calculate the body’s amplitude and initial phase considering the initial conditions given above. [Angular frequency of the particle is π/ s]

Answer:

Given,

Initially, at t = 0:

Displacement, x = 1 cm

Initial velocity, v = ω cm/sec.

Angular frequency, ω = π rad/s

It is given that:

x(t) = A cos( ωt + Φ) . . . . . . . . . . . . . . . . ( i )

1 = A cos( ω x 0 + Φ) = Acos Φ

A cos Φ = 1 . . . . . . . . . . . . . . . . ( ii )

Velocity, v = dx / dt

differentiating equation ( i ) w.r.t ‘t’

v = – Aωsin ( ωt + Φ)

Now at t = 0; v = ω and

=> ω = – Aωsin ( ωt + Φ)

1 = – A sin( ω x 0 + Φ) = -Asin(Φ)

Asin(Φ) = – 1 . . . . . . . . . . . . . . . . . . . . . . . ( iii )

Adding and squaring equations ( ii ) and ( iii ), we get:

A2(sin2 Φ + cos2 Φ) = 1 +1

thus, A =√2

Dividing equation ( iii ) by ( ii ), we get :

tan Φ = -1

Thus, Φ =3π/4 , 7π/4

Now if simple harmonic motion is given as :

x = B sin( ωt + α)

Putting the given values in the equation , we get :

1 = B sin ( ω x 0 + α)

Bsin α = 1 . . . . . . . . . . . . . . . . . . . . ( iv )

Also, velocity ( v ) = ω Bcos (ωt + α)

Substituting the values we get :

π = π B sin α

 B sin α = 1 . . . . . . . . . . . . . . . . . . . . ( v )

Adding and squaring equations ( iv ) and ( v ), we get:

B2[ sin2 α + cos2 α] =2

Therefore, B = √ 2

Dividing equation ( iv ) by equation ( v ), we get :

B sin α / B cos α = 1

tan α =1 = tan (π/4)

Therefore, α = π/4 , 5π/4, ......


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